4. Graphing Rational Functions
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Chapter 4
4.
Graphing Rational Functions
Rational functions are mathematical expressions that represent the ratio of two polynomials. These functions can exhibit various behaviors on a graph, such as discontinuities, which can be in the form of holes, jumps, or asymptotes. Asymptotes can be vertical, horizontal, or even oblique. The graph of a rational function can provide insights into real-world scenarios, such as modeling the annual income per capita or understanding the dynamics of road maintenance. Additionally, the graph can depict the behavior of medication in the bloodstream over time or the cost dynamics of producing items in a factory. Understanding how to graph these functions is crucial for interpreting and predicting real-world phenomena.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Graphing Rational Functions
Slide of 14
A reciprocal function can be considered as the quotient where the denominator is a polynomial with a degree of 1. In other words, there is a variable in the denominator. In this lesson, rational functions in which the numerator and the denominator have a degree of at least 1 will be graphed.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
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Average Cost Function
LaShay's school, Jeffereson High, visits a local pencil factory. The average cost, in thousand dollars, of producing x thousand pencils is given by the following function. A(x)= 0.2x^2+50x+500/x
a Graph the average cost function.
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Rational Functions
A rational function is a function that contains a rational expression. Any function that can be written as the quotient of two polynomial functions p(x) and q(x) is a rational function.
f(x) = p(x)/q(x), q(x) ≠ 0
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Discontinuity of Rational Functions
The graph of a rational function can be a smooth continuous curve, or it can have jumps, breaks, or holes. By looking at the graph of a function, functions can be categorized into two groups: continuous or discontinuous.
Concept
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Discontinuous - Function
A function is said to be continuous if its graph can be drawn without lifting the pencil. Otherwise, the function is said to be discontinuous. 
A discontinuous function can have holes, jumps, or asymptotes throughout its graph. 
A point where a function is discontinuous is called a discontinuity. Function I has a point of discontinuity at x=1, Function II has a point of discontinuity at x=2, and Function III has a point of discontinuity at x=0.
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Point of Discontinuity
A point of discontinuity of a function is a point with the x-coordinate that makes the function value undefined. A point of discontinuity can also be considered as an excluded value of the function rule.
Removable Discontinuity
If a function can be redefined so that its point of discontinuity is removed, it is called a removable discontinuity. A removable discontinuity may occur when there is a rational expression with common factors in the numerator and denominator. Consider the following rational function. f(x) = x^2-4/x-2 For this function, x=2 is a point of discontinuity because its denominator is 0 when x=2. Next, notice that the function can be simplified using a difference of squares.f(x) = x^2-4/x-2
▼
Simplify right-hand side
WritePow
Write as a power
f(x) = x^2-2^2/x-2
FacDiffSquares
a^2-b^2=(a+b)(a-b)
f(x) = (x+2)(x-2)/x-2
SimpQuot
Simplify quotient
f(x) = x + 2, x≠ 2
However, the function can be made continuous by redefining it at x=2. If 2 is substituted for x in the function f(x)=x+2, it is found that f(2)=4. Therefore, the following function is continuous at x=2.
g(x) = x^2-4/x-2, & if x ≠ 2 [0.7em]
4, & if x = 2
This means that x=2 is a removable discontinuity. Note that a removable discontinuity is typically a hole
in the graph.
Non-Removable Discontinuity
When a function cannot be redefined so that the point of discontinuity becomes a valid input, it is called a non-removable discontinuity. Consider, for example, y = 1x+1. Since x=-1 is not in the domain of the function, there is a point of discontinuity at x=- 1.
Example
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Modeling Annual Income Per Capita
LaShay lives in Des Moines, where the total annual amount of personal income I in millions dollars, and the population P in millions, are modeled by the following functions. I(t) & = 5000t+100 000/0.02t+1 [0.8em] P(t) & = 0.36t +10 800 In these functions, t represents the number of years that have passed since 1990.
a Write a function C(t) modeling the annual income per capita in Des Moines.
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b Identify the points of discontinuity of C(t), if any.
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Hint
a Divide the total amount of personal income by the population. Recall how to divide rational expressions.
b For which values of t is the function C(t) undefined?
Solution
a Take a look at what the given equations represent.
| Total Annual Income | Population |
|---|---|
| I(t) = 5000t+100 000/0.02t+1 | P(t) = 0.36t +10 800 |
C(t)= I(t) ÷ P(t)
SubstituteExpressions
Substitute expressions
C(t)= (5000t+100 000/0.02t+1) ÷ (0.36t+10 800)
DivFracSL
.a/b /c.= a/b* c
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800)
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800)
MultPar
Multiply parentheses
C(t)= 5000t+100 000/0.0072t^2+216t+0.36t+10 800
AddTerms
Add terms
C(t)= 5000t+100 000/0.0072t^2+216.36t+10 800
b Consider the function C(t) written in Part A.
C(t)= 5000t+100 000/(0.02t+1)(0.36t+10 800) Recall that division by zero is not defined. Therefore, the rational function is undefined when either of the parentheses equals zero.
| C(t) is undefined | |
|---|---|
| 0.02t+1=0 | 0.36t+10 800=0 |
| t= - 50 | t= - 30 000 |
This means that t=- 50 and t=- 30 000 are excluded values. Domain All real numbers except t= - 50 and t= - 30 000 If a real number a is not in the domain of a function, then the function has a point of discontinuity at x=a. Since the domain is all real numbers except t=- 50 and t=- 30 000, there are two points of discontinuity of the function. Points of Discontinuity t= - 50 and t= - 30 000
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Finding Point of Discontinuity
On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole
on the road.
The equation below models the road to the factory. R(x) = 3x^2+3x/x^3+x^2+x+1
a Find the x-coordinate of the point where the road maintenance is being completed.
b The municipality of Des Moines plans to build a new road, which is modeled by the following function.
Answer
a x=- 1
b Graph:
Hint
a Start by factoring the numerator and the denominator of the function. Find the point of discontinuity of the graph.
b Check if the numerator and the denominator have a common factor.
Solution
a The terms in the numerator of the function have a common factor of 3x. Also, the denominator can be factored by grouping.
R(x) = 3x^2+3x/x^3+x^2+x+1
FactorOut
Factor out 3x
R(x) = 3x(x+1)/x^3+x^2+x+1
FactorOut
Factor out x^2
R(x) = 3x(x+1)/x^2(x+1)+x+1
FactorOut
Factor out (x+1)
R(x) = 3x(x+1)/(x^2+1)(x+1)
Notice that it is also a removable discontinuity because x+1 is the common factor in the numerator and denominator.
b Start by factoring the numerator and denominator to see if the function can be simplified.
N(x) = 2x^2-x-3/10x+10
N(x) = (2x-3)(x+1)/10(x+1)
This graph also has a removable discontinuity at x=- 1 because x+1 is the common factor in the numerator and denominator of the rational function.
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Asymptotes of Rational Functions
An asymptote of a graph is an imaginary line that the graph gets close to as x goes to plus or minus infinity or a particular number. For example, the graph of the rational function f(x) = 1x has two asymptotes — the x-axis and the y-axis.
Analyzing the diagram, the following can be observed.
- As x approaches infinity and as x approaches negative infinity, the value of the function approaches 0. Therefore, y = 0, or the x-axis, is a horizontal asymptote of the graph of f.
- As x approaches 0, the value of the function approaches either positive or negative infinity. Therefore, x = 0, or the y-axis, is a vertical asymptote of the graph of f.
Consider a rational function where p(x) is a polynomial with a leading coefficient of a and a degree of m and where q(x) is a polynomial with a leading coefficient of b and a degree of n. f(x) = p(x)/q(x) = ax^m + .../bx^n + ... To determine the asymptotes of the function, the denominator and the degrees of the polynomials in the numerator and denominator must be analyzed.
Identifying Vertical Asymptotes
To identify the vertical asymptotes, the function should be in its simplest form. That is, if p(x) and q(x) have common factors, the function should be simplified first. The vertical asymptotes will occur at the zeros of the denominator. Check out the following examples.
Identifying Horizontal Asymptotes
The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.
| f(x) = p(x)/q(x) = ax^m + .../bx^n + ... | |
|---|---|
| Horizontal Asymptote | If m >n, there is no horizontal asymptote. |
| If m < n, the horizontal asymptote is y=0. | |
| If m = n, the horizontal asymptote is y = ab. |
Example
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Identifying Asymptotes
LaShay takes a travel-sickness pill about 45 minutes before the school bus leaves for the field trip. The amount A in milligrams of the medication in her bloodstream is modeled by the following rational function.
A(t) = 6t^2-24t/t^3-t^2-10t-8
Here, t represents the time in hours after one pill is taken. Identify the asymptotes of the function.
Answer
Vertical Asymptotes: t=- 2 and t=- 1
Horizontal Asymptote: y=0
Hint
Factor the numerator. What does the Factor Theorem state?
Solution
To identify the asymptotes, the numerator and denominator of the function A(t) need to be factored. Notice that the terms in the numerator have a common factor of 6t.
To factor the polynomial in the denominator, use the Factor Theorem.
Three roots for p(t) were found, so there is no need for further investigation. In conclusion, - 1, - 2, and 4 are all the real roots of t^3-t^2-10t-8. Next, apply the Factor Theorem to write its factored form.
t^3-t^2-10t-8 = (t+1)(t+2)(t-4)
Now that the factored form of the denominator has been found, the given function can be simplified.
Since the numerator and the denominator share the factor t-4, there is a hole at t=4. Also, the denominator becomes zero when t=- 1 and t=- 2. These values make the function undefined. Therefore, the function has two vertical asymptotes.
Vertical Asymptotes t = - 1 and t = - 2
To find the horizontal asymptote, compare the degrees of the polynomials in the numerator and denominator.
|
Factor Theorem |
|
The expression x-a is a factor of a polynomial if and only if the value a is a zero of the related polynomial function. |
Now, the aim is to find all real zeros of the polynomial. Before doing so, identify the degree of the polynomial. p(t)=t^3-t^2-10t-8 It is a polynomial with a degree of 3. Subsequently, by the Fundamental Theorem of Algebra, it is known know that p(t) has exactly three roots. Additionally, by the Rational Root Theorem, integer roots must be factors of the constant term. Since the constant term of p(t) is - 8, the possible integer roots are ± 1, ± 2, ± 4, and ± 8.
| t | t^3-t^2-10t-8 | p(t)=t^3-t^2-10t-8 |
|---|---|---|
| 1 | 1^3-( 1)^2-10( 1)-8 | - 18 |
| - 1 | - 1^3-( - 1)^2-10( - 1)-8 | 0 |
| 2 | 2^3-( 2)^2-10( 2)-8 | - 24 |
| - 2 | - 2^3-( - 2)^2-10( - 2)-8 | 0 |
| 4 | 4^3-6( 4)^2-7( 4)+60 | 0 |
A(t) = 6t(t-4)/t^3-t^2-10t-8
Substitute
t^3-t^2-10t-8= (t+1)(t+2)(t-4)
A(t) = 6t(t-4)/(t+1)(t+2)(t-4)
CrossCommonFac
Cross out common factors
A(t) = 6t(t-4)/(t+1)(t+2)(t-4)
CancelCommonFac
Cancel out common factors
A(t) = 6t/(t+1)(t+2)
| y=ax^m/bx^n | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
Look at the degrees of the numerator and denominator of the function. A(t) = 6t^2-24t/t^3-t^2-10t-8 The degree of the denominator is greater than the degree of the numerator, m < n. Therefore, the horizontal asymptote is y=0. Horizontal Asymptotes y =0
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Oblique Asymptote
An asymptote can be neither vertical nor horizontal. It can be a slanted line. In such cases, it is called an oblique asymptote or slant asymptote.
An oblique asymptote of a rational function occurs when the degree of the numerator is 1 more than that of the denominator. Therefore, a function with an oblique asymptote can never have a horizontal asymptote. To find the equation of the oblique asymptote of a rational function f(x)= p(x)q(x), p(x) is divided by q(x) using long division to get a quotient ax+b with a remainder r(x). f(x) = p(x)/q(x) ⇓ f(x) = ax+b+r(x) The equation of the oblique asymptote is of the form y = ax + b, where a is a non-zero real number.
Example
To illustrate how the technique is used, the asymptotes of the following function will be found. f(x) = x^3+2x/4x^2-1 The difference between the degrees of the numerator and denominator is 1. This means that its graph does not have a horizontal asymptote but an oblique asymptote. To find the oblique asymptote, polynomial long division will be used.l r 4x^2 - 1 & |l x^3 + 2x
▼
Divide
x^3/4x^2= x/4
r x4 r 4x^2-1 & |l x^3+2x
Multiply term by divisor
r x4 rl 4x^2 - 1 & |l x^3+2x & x^3- x4
Subtract down
r x4 r 4x^2 - 1 & |l 9x4
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Modeling Interest in The Trip
The students spent about 3 hours at the factory. At the end of this period, LaShay's teacher made a survey about the trip and drew a graph relating the students' interest in the trip and the time elapsed at the factory. The greater the S(t) value, the greater the interest of the students.
Answer
Vertical Asymptotes: t=- 1
Oblique Asymptote: O(t)=- t/2+2
Hint
Start by subtracting the rational expressions on the right-hand side of the function.
Solution
First, the rational expressions on the right-hand side will be subtracted. The least common denominator of the expressions is 2(t+1)^2. Therefore, the first expression needs to be multiplied by 2(t+1)2(t+1) so that both have the same denominator.
The numerator and the denominator do not have any common factors. Therefore, the function does not have a point of discontinuity, or hole, in its graph. However, it has a vertical asymptote at t=- 1 because the function is undefined at that value.
Vertical Asymptote t = -1
Now, to determine if the given rational function has a horizontal or oblique asymptote, the degrees of the polynomials in the numerator and the denominator will be compared.
S(t) = - t^3+2t^2+2t/2(t+1)^2 ⇕ S(t) = - t^3+2t^2+2t/2t^2+4t+2
The degree of the numerator is greater than the degree of the denominator, so the graph has no horizontal asymptotes. However, since the degree of the numerator is 1 more than that of the denominator, the graph has an oblique asymptote. Its equation is found by dividing the numerator by the denominator. Use long division to do this.
The given function can be rewritten as follows.
S(t) = - t/2 + 2 + - 5t-4/2t^2+4t+2
Therefore, the equation of the oblique asymptote is O(t)=- t2+2.
S(t) = t/t+1-t^3/2(t+1)^2
▼
Simplify right-hand side
ExpandFrac
a/b=a * 2(t+1)/b * 2(t+1)
S(t) = t * 2(t+1)/(t+1)* 2(t+1)-t^3/2(t+1)^2
Multiply
Multiply
S(t) = 2t(t+1)/2(t+1)^2-t^3/2(t+1)^2
SubFrac
Subtract fractions
S(t) = 2t(t+1)-t^3/2(t+1)^2
Distr
Distribute 2t
S(t) = 2t^2+2t-t^3/2(t+1)^2
CommutativePropAdd
Commutative Property of Addition
S(t) = - t^3+2t^2+2t/2(t+1)^2
▼
Rewrite the denominator
S(t) = - t^3+2t^2+2t/2t^2+4t+2
l r 2t^2+4t+2 & |l - t^3+2t^2+2t
▼
Divide
- t^3/2t^2= - t/2
r - t2 r 2t^2+4t+2 & |l - t^3+2t^2+2t
Multiply term by divisor
r - t2 rl 2t^2+4t+2 & |l - t^3+2t^2+2t & - t^3 -2t^2-t
Subtract down
r - t2 r 2t^2+4t+2 & |l 4t^2+3t
▼
Divide
4t^2/2t^2= 2
r - t2 + 2 r 2t^2+4t+2 & |l 4t^2+3t
Multiply term by divisor
r- t2 + 2 rl 2t^2+4t+2 & |l 4t^2+3t & 4t^2+8t+4
Subtract down
r - t2 + 2 r 2t^2+4t+2 & |l - 5t-4
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Graphing Rational Functions
Rational functions may have more than one vertical asymptote, which can make graphing them a bit more difficult than graphing reciprocal functions. However, the asymptotes are useful when graphing a rational function because they can describe the end behavior. Consider graphing the following function.
f(x) =(x^2+5x-6)(x-3)/x^2-7x+12
To graph the rational function, these four steps can be followed.
1
Draw the Asymptotes
expand_more First, the vertical asymptote(s) of the function will be found. If the numerator and the denominator of a rational function have no common factors, then the vertical asymptote is the x-value that makes the denominator zero. Start by factoring the denominator to see candidates for asymptotes.
The rational function is undefined where x-4=0 and x-3=0. This means that x= 4 and x=3 are not included in the domain. Notice that x-3 is a common factor between the numerator and the denominator. Next, check if there are any other common factors and cancel them.
Since the factor x-3 was canceled, the graph has a
Now consider the degrees of the numerator and denominator for the given function. The simplified form of the function can also be considered at this point because both have the same graph.
f(x) =(x-1)(x+6)/x-4, x≠ 3 ⇓ f(x) =x^2+5x-6/x^1-4, x ≠ 3
The degree of the numerator minus the degree of the denominator equals 1. Therefore, the function has an oblique asymptote. The equation of the oblique asymptote is found by dividing the numerator by the denominator without considering the remainder. To divide the polynomials, long division will be used.
The equation of the oblique asymptote is y=x+9. Finally, draw the asymptotes on a coordinate plane. 
f(x) =(x^2+5x-6)(x-3)/x^2-7x+12
f(x) =(x^2+5x-6)(x-3)/(x-4)(x-3)
f(x) =(x^2+5x-6)(x-3)/(x-4)(x-3)
f(x) =(x-1)(x+6)(x-3)/(x-4)(x-3)
▼
Simplify
CrossCommonFac
Cross out common factors
f(x) =(x-1)(x+6)(x-3)/(x-4)(x-3)
CancelCommonFac
Cancel out common factors
f(x) =(x-1)(x+6)/x-4, x ≠ 3
holeat x=3. Also, the graph has a vertical asymptote at x=4 because 4 is not included in the domain. Hole: & x =3 Vertical Asymptote: & x = 4 The degrees of the polynomials in the numerator and denominator determine whether the function has a horizontal or oblique asymptote. In the following table, m and n are the degrees of the polynomials in the numerator and denominator, and a and b are their respective leading coefficients.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
l r x-4 & |l x^2+5x-6
▼
Divide
x^2/x= x
r x r x-4 & |l x^2+5x-6
Multiply term by divisor
r x rl x-4 & |l x^2+5x-6 & x^2-4x
Subtract down
r x r x-4 & |l 9x-6
▼
Divide
9x/x= 9
r x+9 r x-4 & |l 9x-6
Multiply term by divisor
rx + 9 rl x-4 & |l 9x - 6 & 9x-36
Subtract down
r x+9 r x-4 & |l 30
The given function has one vertical asymptote. Therefore, its graph will consist of two parts, one to the left and the other to the right of the vertical asymptote. These parts may lie above or below the oblique asymptote.
2
Find the Intercepts
expand_more The x-intercepts are the points where the graph intersects the x-axis. The value of the y-coordinate is zero at these points. Substitute 0 for f(x) in the given function and solve for x. The simplified function can be used to find the zeros.
The graph has two x-intercepts. The y-intercept is the point where the graph intersects the y-axis. At this point, the value of the x-coordinate is zero.
There is a y-intercept at (0,1.5). Show the intercepts on the same coordinate plane. 
f(x) =(x-1)(x+6)/x-4
Substitute
f(x)= 0
0=(x-1)(x+6)/x-4
▼
Solve for x
MultEqn
LHS * (x-4)=RHS* (x-4)
0=(x-1)(x+6)
ZeroProdProp
Use the Zero Product Property
lcx-1=0 & (I) x+6=0 & (II)
AddEqn
(I): LHS+1=RHS+1
lx=1 x+6 = 0
SubEqn
(II): LHS-6=RHS-6
lx=1 x=- 6
3
Make a Table of Values
expand_more Some additional points are needed to get a good sense of the shape of the graph. Make sure to only use values included in the domain of the function.
| x | (x-1)(x+6)/x-4 | f(x)=(x-1)(x+6)/x-4 |
|---|---|---|
| - 11 | ( - 11-1)( - 11+6)/- 11-4 | - 4 |
| 2 | ( 2-1)( 2+6)/2-4 | - 4 |
| 6 | ( 6-1)( 6+6)/6-4 | 30 |
| 9 | ( 9-1)( 9+6)/9-4 | 24 |
| 19 | ( 19-1)( 19+6)/19-4 | 30 |
It is almost done. Plot the points and imagine how the shape of the graph should look!
As can be seen, one part of the graph will lie to the left of the vertical asymptote and below the oblique asymptote. The other part of the graph will be to the right of the vertical asymptote and above the oblique asymptote.
4
Draw the Graph
expand_more The graph can now be drawn by connecting the points with a smooth curve. It must approach the asymptotes. Do not forget to plot the hole at x=3!
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Graphing the Function for the Surface Area to Volume Ratio of a Box
The factory sells 6 pencils in a rectangular box. The dimensions of the box are shown in the diagram.
a Write a function of the form E(x)= S(x)V(x), where S(x) and V(x) are the surface area and volume of the box, respectively.
b Graph the function E(x).
Answer
a E(x) = x^2+4x-2/x^2+x-6
b Graph:
Hint
a The surface area of a prism is the sum of the areas of its surfaces. The volume of a prism is the product of its dimensions.
b Start by determining the vertical and horizontal asymptotes. Then, make a table of values.
Solution
a First, the surface area and volume of the given box will be found.
S(x)= 2( 2)( x-2) + 2( 2x+6)( x-2)+2( 2)( 2x+6)
▼
Simplify right-hand side
Multiply
Multiply
S(x)= 4(x-2) + 2(2x+6)(x-2)+4(2x+6)
Distr
Distribute 4
S(x)= 4x-8 + 2(2x+6)(x-2)+8x+24
Distr
Distribute 2
S(x)= 4x-8 + (4x+12)(x-2)+8x+24
MultPar
Multiply parentheses
S(x)= 4x-8 +4x^2-8x+12x-24+8x+24
AddSubTerms
Add and subtract terms
S(x)= 4x^2+16x-8
V(x) = ( 2x+6)( x-2)( 2)
V(x) = 4x^2+4x-24
E(x) = S(x)/V(x)
SubstituteExpressions
Substitute expressions
E(x) = 4x^2+16x-8/4x^2+4x-24
▼
Simplify right-hand side
E(x) = x^2+4x-2/x^2+x-6
b To graph the given rational function, its asymptotes and intercepts will be found. A table of values will then be used to draw the graph.
Asymptotes
Consider the function written in the previous part. E(x)=x^2+4x-2/x^2+x-6 First, factor the denominator of this function to identify its domain.E(x)=x^2+4x-2/x^2+x-6
WriteDiff
Write as a difference
E(x)=x^2+4x-2/x^2+3x-2x-6
E(x)=x^2+4x-2/x(x+3)-2(x+3)
FactorOut
Factor out (x+3)
E(x)=x^2+4x-2/(x-2)(x+3)
holesin the graph. Recall that if the real number a is not included in the domain, there is a vertical asymptote at x=a. In this case, there are two vertical asymptotes, one at x=- 3 and the other at x=2. Vertical Asymptotes x=- 3 and x =2 To find whether the graph has a horizontal or oblique asymptote, review how its type is determined using the degrees of the polynomials in the numerator and the denominator.
| y=ax^m+.../bx^n+... | Asymptote |
|---|---|
| m | y=0 |
| m>n | none |
| m=n | y=a/b |
| m-n=1 | y= the quotient of the polynomials with no remainder |
Look at the degrees of the numerator and denominator of the function. E(x)=1x^2+4x-2/1x^2+x-6 The degree of the denominator is equal to the degree of the numerator. Therefore, the line y= 1 1=1 is the horizontal asymptote of the function.
Intercepts
To find the x-intercepts, 0 is substituted for E(x) into the function. It will be better to use the form in which the numerator and the denominator are factored.E(x) = (x+2+sqrt(6))(x+2-sqrt(6))/(x-2)(x+3)
Substitute
E(x)= 0
0 = (x+2+sqrt(6))(x+2-sqrt(6))/(x-2)(x+3)
▼
Solve for x
MultEqn
LHS * (x-2)(x+3)=RHS* (x-2)(x+3)
0 = (x+2+sqrt(6))(x+2-sqrt(6))
ZeroProdProp
Use the Zero Product Property
lcx+2+sqrt(6)=0 & (I) x+2-sqrt(6)=0 & (II)
SubEqn
(I): LHS-2-sqrt(6)=RHS-2-sqrt(6)
lx=- 2- sqrt(6) x+2-sqrt(6)=0
SubEqn
(II): LHS-2+sqrt(6)=RHS-2+sqrt(6)
lx=- 2 - sqrt(6) x = - 2+sqrt(6)
UseCalc
Use a calculator
lx=- 4.449489 ... x = 0.449489 ...
RoundDec
Round to 1 decimal place(s)
lx ≈ - 4.4 x ≈ 0.4
E(x)=x^2+4x-2/x^2+x-6
Substitute
x= 0
E( 0)=0^2+4( 0)-2/0^2+ 0-6
▼
Solve for E(0)
CalcPow
Calculate power
E(0)=0+4* 0-2/0+0-6
ZeroPropMult
Zero Property of Multiplication
E(0)=0+0-2/0+0-6
IdPropAdd
Identity Property of Addition
y=- 2/- 6
ReduceFrac
a/b=.a /(- 2)./.b /(- 2).
y=1/3
Table of Values
Now, make a table of values to graph the given function.
| x | x^2+4x-2/x^2+x-6 | E(x)=x^2+4x-2/x^2+x-6 |
|---|---|---|
| - 6 | ( - 6)^2+4( - 6)-2/( - 6)^2+( - 6)-6 | ≈ 0.417 |
| - 3.5 | ( - 3.5)^2+4( - 3.5)-2/( - 3.5)^2+( - 3.5)-6 | ≈ - 1.364 |
| - 2 | ( - 2)^2+4( - 2)-2/( - 2)^2+( - 2)-6 | 1.5 |
| - 1 | ( - 1)^2+4( - 1)-2/( - 1)^2+( - 1)-6 | ≈ 0.833 |
| 1 | ( 1)^2+4( 1)-2/( 1)^2+( 1)-6 | - 0.75 |
| 3 | ( 3)^2+4( 3)-2/( 3)^2+( 3)-6 | ≈ 3.167 |
| 4 | ( 4)^2+4( 4)-2/( 4)^2+( 4)-6 | ≈ 2.143 |
| 5 | ( 5)^2+4( 5)-2/( 5)^2+( 5)-6 | ≈ 1.792 |
Finally, the graph of the function can be drawn by plotting the found points and connecting them with a smooth curve. Recall that a rational function can cross the horizontal asymptote but cannot cross the vertical asymptotes. In this case, the horizontal asymptote is crossed.
Extra
Factoring The Numerator of E(x)Recall the Factor Theorem.
|
p(x) has a factor (x−a) if and only if p(a)=0. |
x = - b ± sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x = - 4 ± sqrt(4^2-4(1)(- 2))/2(1)
▼
Evaluate right-hand side
IdPropMult
Identity Property of Multiplication
x = - 4 ± sqrt(4^2-4(- 2))/2
CalcPow
Calculate power
x = - 4 ± sqrt(16-4(- 2))/2
MultNegNegOnePar
- a(- b)=a* b
x = - 4 ± sqrt(16+8)/2
AddTerms
Add terms
x = - 4 ± sqrt(24)/2
SplitIntoFactors
Split into factors
x = - 4 ± sqrt(4 * 6)/2
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x = - 4 ± sqrt(4) * sqrt(6)/2
CalcRoot
Calculate root
x = - 4 ± 2 * sqrt(6)/2
FactorOut
Factor out 2
x = 2( - 2 ± sqrt(6))/2
SimpQuot
Simplify quotient
x = - 2 ± sqrt(6)
Example
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Modeling the Amount of Water Pumped Into the Factory
Factory managers plan to pump water into the factory from a pond near the factory. The managers modeled the amount of water pumped in thousands of barrels per year x years after pumping starts to be as follows. W(x) =- x^3+140x^2+1500x+5250/15x^2+150x
a Use a graphing calculator to find the x-intercepts rounded to the nearest integer. Interpret the x-intercepts.
b Determine if the graph has horizontal or oblique asymptotes. If the graph has an oblique asymptote, find its equation.
Answer
a x-intercept: (150,0)
Interpretation: In about 150 years, there will be no water left in the pond.
b The graph has an oblique asymptote because the difference between the degrees of the numerator and the denominator is 1.
Equation: y = - x/15+10
Hint
a Use the zero option in the calculator. It may be necessary to re-size the viewing window in order to see the entire graph.
b Compare the degrees of the numerator and the denominator.
Solution
Now, push GRAPH to draw the function.
It appears that only one part of the graph is shown, but it is not possible to know that until the size of the viewing window is changed.
The graph does not seem to have any zeros. However, note that as x increases in the first quadrant, the graph gets closer to the x-axis and could cross it. To check this, zoom in on this part by changing the window settings. Push WINDOW and change the settings as shown below. Then push GRAPH once more to draw the equation with these new settings.
Now it can be seen that the graph intersects the x-axis and there is a zero. To find it, use the zero option in the calculator. It can be found by pressing 2ND and then CALC.
After selecting the zero
option, choose left and right boundaries for the zero. Finally, the calculator asks for a guess where the zero might be. After that, it will calculate the exact point.
The function intersects the x-axis at about (150,0). Since x represents the numbers of years that have passed since pumping starts, the x-intercept means that in about 150 years, there will be no water left in the pond.
b Recall the table used to determine what type of asymptote a rational function can have.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
l r 15x^2+150x & |l - x^3+140x^2+1500x+5250
▼
Divide
- x^3/15x^2= - x/15
r - x15 r 15x^2+150x & |l - x^3+140x^2+1500x+5250
Multiply term by divisor
r - x15 rl 15x^2+150x & |l - x^3+140x^2+1500x+5250 & - x^3 - 10x^2
Subtract down
r - x15 r 15x^2+150x & |l 150x^2+1500x+5250
▼
Divide
150x^2/15x^2= 10
r - x15 + 10 r 15x^2+150x & |l 150x^2+1500x+5250
Multiply term by divisor
r- x10 + 10 rl 15x^2+150x & |l 150x^2+1500x+5250 & 150x^2+1500x
Subtract down
r - x15 + 10 r 15x^2+150x & |l 5250
Closure
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Graphing the Average Cost Function
Rational functions can have more than one vertical asymptote but at most one horizontal or oblique asymptote. Also, if the numerator and denominator have a common factor (x-a), then there is a hole at x=a. With these tips in mind, the graph of the function that LaShay wrote at the beginning of the lesson can now be graphed. A(x)= 0.2x^2+5x+500/x The function gives the average cost A(x) in thousands dollars for producing x thousand pencils.
a Graph the average cost function.
Answer
a Graph:
b Domain: (- ∞, 0) ⋃ (0,∞)
Range: (- ∞, - 15] ⋃ [25, ∞)
Range: (- ∞, - 15] ⋃ [25, ∞)
Hint
a Start by identifying asymptotes. Then, make a table of values using values that are around the vertical asymptote.
b Examine the graph and see if the graph is symmetric about the point of intersection of the asymptotes. Determine all the y-values for which the function is graphed.
Solution
a The steps to draw a rational function can be listed as follows.
- Draw the asymptotes.
- Find any intercepts.
- Make a table of values.
- Draw the graph.
Drawing the Asymptotes
The vertical asymptotes of a rational function are the x-values where the denominator of the function is zero. A(x) = 0.2x^2+5x+500/x The denominator is 0 when x=0. This means that this value is not included in the domain. Domain All real numbers except x =0 Since the numerator and denominator do not share a common factor, the graph does not contain any holes. Additionally, the graph has a vertical asymptote at the x-value excluded from the domain. Vertical Asymptote x =0 Next, recall the table for determining the other asymptotes.
| y=ax^m+.../bx^n+... | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | None | None |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
l r x & |l 0.2x^2+5x+500
▼
Divide
0.2x^2/x= 0.2x
r 0.2x r x & |l 0.2x^2+5x+500
Multiply term by divisor
r 0.2x rl x & |l 0.2x^2+5x+500 & 0.2x^2
Subtract down
r 0.2x r x & |l 5x+500
▼
Divide
5x/x= 5
r 0.2x + 5 r x & |l 5x+500
Multiply term by divisor
r0.2x + 5 rl x & |l 5x+500 & 5x
Subtract down
r 0.2x+5 r x & |l 500
Finding the Intercepts
The x-intercept and y-intercept are the points where a graph crosses the x- and y-axes, respectively. Subsequently, the x-intercepts occurs when A(x)=0.A(x) =0.2x^2+5x+500/x
Substitute
A(x)= 0
0=0.2x^2+5x+500/x
MultEqn
LHS * x=RHS* x
0= 0.2x^2+5x+500
▼
Solve using the quadratic formula
UseQuadForm
Use the Quadratic Formula: a = 0.2, b= 5, c= 500
x=- 5±sqrt(5^2 -4( 0.2)( 500))/2( 0.2)
CalcPowProd
Calculate power and product
x=- 5±sqrt(25 -400)/0.4
SubTerm
Subtract term
x=- 5±sqrt(- 375)/0.4
Making a Table of Values
A table of values will be used to get a rough idea of the shape of the graph.
| x | 0.2x^2+5x+500/x | A(x)=0.2x^2+5x+500/x |
|---|---|---|
| - 75 | 0.2( - 75)^2+5( - 75)+500/- 75 | ≈ - 16.7 |
| - 25 | 0.2( - 25)^2+5( - 25)+500/- 25 | - 20 |
| - 10 | 0.2( - 10)^2+5( - 10)+500/- 10 | - 47 |
| 10 | 0.2( 10)^2+5( 10)+500/10 | 57 |
| 25 | 0.2( 25)^2+5( 25)+500/25 | 30 |
| 75 | 0.2( 75)^2+5( 75)+500/75 | ≈ 26.7 |
Plot the points and imagine how the shape of the graph should look!
Finally, draw the graph by connecting the points. It must approach, but not cross, the asymptotes.
b The domain of the function was identified when graphing the function.
Domain: (- ∞,0) ⋃ (0,∞) To identify the range of the function, take a look at its graph.
As shown, the range does not contain all real numbers. It appears that the minimum point of the part of the graph in the first quadrant is (50,25). Notice also that the graph is symmetric about the point of intersection of the asymptotes, (0,5). Using this fact, the maximum point of the other part can be identified.
It is the point (- 50,- 15). Therefore, the range is all the y-values less than or equal to - 15 and greater than or equal to 25. Range: (∞,- 15] ⋃ [25,∞)
Checking Our Answer
Use a Graphing CalculatorA graphing calculator can be used to check the answers. The graph of the function will be drawn first. Press the Y= button and type the equation in the first row.
By pushing GRAPH, the calculator will draw the equation. For this function to be visible on the screen, re-size the standard window by pushing the WINDOW button. Change the settings to a more appropriate size and then push GRAPH.
Next, the local maximum and local minimum will be found. To do so, push 2nd, then TRACE, and choose the maximum
option.
When using the maximum
feature, choose the left and right bounds. The calculator will then provide a best guess as to where the maximum might be.
The maximum point of the graph in the third quadrant is (- 50,- 15). To find the minimum, repeat the same process but choose the minimum
option.
The minimum point of the part in the first quadrant is (50,25). As a result, the range is (∞,- 15] ⋃ [25,∞), and the domain is all real numbers except 0.
Graphing Rational Functions
Exercise 2.1
Diego lists some of the characteristics of the graph of the function y = x^2-7x+12x^2+8x+15 as follows.
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We will start by factoring the numerator and denominator of the given rational function.
We can see that the rational function is undefined when x+5=0 and x+3=0. This means that x=- 5 and x=- 3 are not included in the domain. Notice also that the numerator and denominator do not have any common factors. Therefore, we can conclude that the graph of the function has vertical asymptotes at x=- 5 and x=- 3, but no holes. Vertical Asymptotes x = - 5 and x = - 3 To determine if the function has an oblique or horizontal asymptote, we first need to compare the degrees of the polynomials in the numerator and denominator of the rational function.
| y=ax^m/bx^n | Asymptote | Asymptote Type |
|---|---|---|
| m | y=0 | Horizontal |
| m>n | none | none |
| m=n | y=a/b | Horizontal |
| m-n=1 | y= the quotient of the polynomials with no remainder | Oblique |
In this case, we have a rational function whose numerator and denominator have a degree of 2. y=x^2-7x+12/x^2+8x+15 Therefore, the function has a horizontal asymptote, but not an oblique asymptote. The horizontal asymptote is the ratio of the coefficient of the term with a greatest degree in the numerator to the coefficient of the term with a greatest degree in the denominator. y=( 1)x^2-7x+12/( 1)x^2+8x+15 ⇒ y=1/1=1 Therefore, the horizontal asymptote is y=1. Horizontal Asymptote y = 1 Diego thinks that the zeros of the numerator are the vertical asymptotes and that the zeros of the denominator are the horizontal asymptotes. However, we have shown that this is not the case. As a result, of the given statements, only IV is correct.
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Vincenzo plays basketball for South High School. He has managed to make 14 out of 25 free throw attempts so far this season. He is determined to increase his rate of free throws made — also referred to as a free throw percentage.
If he makes x free throw attempts in a row, his rate can be calculated using the function R(x).
R(x) =14+x/25+x
Which graph represents the function R(x)?
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To graph the given rational function we need to find its domain, asymptotes, and intercepts. Next, we can create a table of values by evaluating the function rule. Finally, we will plot and connect those points.
| R(x) = 14+x/25+x | Explanation | |
|---|---|---|
| Domain | All real numbers except x = -25 | When x = - 25, the function is undefined. |
| Vertical Asymptote | x=- 25 | The fraction is in simplest form and - 25 is not included in the domain. |
| Horizontal Asymptote | y= 1/1 = 1 | The ratio of the coefficients of the x-variables in the numerator and denominator. |
| x-intercept | (- 14,0) | The solution of 14+x25+x=0 is x = - 14. |
| y-intercept | (0,0.56) | The value of 14+025+0 is 0.56. |
Now, to find points on the curve, we can make a table of values. Make sure to include values for x to the left and to the right of the vertical asymptote.
| x | 14+x/25+x | R(x) |
|---|---|---|
| - 35 | 14+( - 35)/25+( - 35) | 2.1 |
| - 30 | 14+( - 30)/25+( - 30) | 3.2 |
| - 26 | 14+( - 26)/25+( - 26) | 12 |
| - 24 | 14+( - 24)/25+( - 24) | - 10 |
| - 7 | 14+( - 7)/25+( - 7) | ≈ 0.4 |
| 10 | 14+( 10)/25+( 10) | ≈ 0.7 |
Let's plot and connect the obtained points. Keep in mind that rational functions have two branches. Remember to graph the asymptotes!
Given that x is the number of consecutive free throw attempts, we can assume that x is a non-negative number. Therefore, the part in the first quadrant is meaningful.
This graph matches with graph B.
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Dominika drew the graphs of the following functions.
f(x) & = x-4 [0.8em]
g(x) & = (x+6)(x-4)/x+6
Which of the following statements could Dominika have said after looking at the graphs she drew?A. & Both graphs are exactly the same, except
& that the graph of g(x) has a hole atx=- 4. [0.5em]
B. & Both graphs are exactly the same, except
& that the graph of f(x) has a hole atx=- 6. [0.5em]
C. &Both graphs are exactly the same. [0.5em]
D. &Both graphs are exactly the same, except
& that the graph of g(x) has a hole atx=-6.
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We are given two functions and asked to identify which of the given statements is right about their graphs of the functions. Let's look at the given functions. f(x) & = x-4 [0.8em] g(x) & = (x+6)(x-4)/x+6 Note that the function f(x) is a linear function. Its graph is a line with a slope of 1 and a y-intercept at (0,-2). The function g(x) is a rational function, which is not defined at x=-6. Because the numerator and denominator have a common factor, we can simplify it. g(x) = (x+6)(x-4)/x+6 ⇓ g(x) = x-4, x≠ -6 After simplifying, we can say that f(x)= g(x) except when x=-6. This means that their graphs are exactly the same, except that the graph of g(x) has a hole at x=-6. That is, g(x) has a point of discontinuity at x=-6.
Therefore, from the given statements, option D is the right one.
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Graphing Rational Functions
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